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Question Number 43659 by Joel578 last updated on 13/Sep/18

$$\mathrm{An}\mathrm{unfair}\mathrm{coin}\mathrm{with}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}$$$$\mathrm{head}\mathrm{in}\mathrm{one}\mathrm{toss}=\frac{1}{5}.$$$$\mathrm{If}\mathrm{coin}\mathrm{tosses}n\mathrm{times},\mathrm{the}\mathrm{probability}\mathrm{of}$$$$\mathrm{getting}2\mathrm{heads}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{probability}\mathrm{of}$$$$\mathrm{getting}3\mathrm{heads}.\mathrm{Find}n$$

Commented by math1967 last updated on 14/Sep/18

$$n=14?$$

Commented by Joel578 last updated on 17/Sep/18

$$\mathrm{Please}\mathrm{explain}\mathrm{the}\mathrm{working}\mathrm{Sir}$$

Answered by math1967 last updated on 17/Sep/18

$$probabilitygettinghead=\frac{1}{5}$$$$probabilityofgettingtail=(1-\frac{1}{5})=\frac{4}{5}$$$$fromtheproblemofbinomialdistribution$$$${\stackrel{n}{c}}_2{\left(\frac{1}{5}\right)}^2\times {\left(\frac{4}{5}\right)}^{n-2}=\stackrel{n}{}{c}_3{\left(\frac{1}{5}\right)}^3\times {\left(\frac{4}{5}\right)}^{n-3}$$$$\frac{n!}{2!(n-2)!}\times {\left(\frac{4}{5}\right)}^{n-2-n+3}=\frac{n!}{3!(n-3)!}\times {\left(\frac{1}{5}\right)}^{3-2}$$$$\frac{4}{10(n-2)}=\frac{1}{30}\Rightarrow 10n=120+20$$$$\therefore n=14$$

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